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Successive numbers n such that ChebyshevT[n/2, n] is not an integer.
1

%I #22 May 01 2015 10:57:32

%S 3,5,9,11,13,15,19,21,23,25,27,29,33,35,37,39,41,43,45,47,51,53,55,57,

%T 59,61,63,65,67,69,73,75,77,79,81,83,85,87,89,91,93,95,99,101,103,105,

%U 107,109,111,113,115,117,119,121,123,125,129,131,133,135,137

%N Successive numbers n such that ChebyshevT[n/2, n] is not an integer.

%C Successive numbers n such that ChebyshevT[x, x/2] is not integer (is integer/2) see A007310

%C From _Robert Israel_, Apr 30 2015 (Start)

%C Numbers 2*k+1 where (1/2)*(2*k+1+2*sqrt(k*(k+1)))^(k+1/2)+(1/2)*(2*k+1+2*sqrt(k*(k+1)))^(-k-1/2) is not an integer.

%C All odd numbers not in A056220. (End)

%C From _Robert Israel_, May 01 2015 (Start)

%C ChebyshevT(1/2,n) = sqrt((n+1)/2) is an integer iff n is in A056220, otherwise it is irrational.

%C For odd k, ChebyshevT(k,x) = x*P(x^2) where P is a polynomial with integer coefficients and no roots >= 1.

%C Therefore if k is odd and n is a positive integer, ChebyshevT(k/2,n) = ChebyshevT(k,ChebyshevT(1/2,n)) = sqrt((n+1)/2)*P((n+1)/2) is an integer iff n is in A056220, otherwise it is irrational. (End)

%H Robert Israel, <a href="/A173263/b173263.txt">Table of n, a(n) for n = 1..1000</a>

%p remove(t -> issqr((t+1)/2), [seq(2*j+1,j=0..1000)]); # _Robert Israel_, Apr 30 2015

%t aa = {}; Do[If[PossibleZeroQ[Round[N[ChebyshevT[x/2, x], 100]] - N[ChebyshevT[x/2, x], 100]], , AppendTo[aa, x]], {x, 0, 1500}]; aa (* _Artur Jasinski_, Feb 14 2010 *)

%Y Cf. A007310, A056220.

%K nonn

%O 1,1

%A _Artur Jasinski_, Feb 14 2010

%E Corrected and edited by _Robert Israel_, Apr 30 2015